Question

struct myType {
    vector<char*> ls;
};

Here ls is holding pointers to char. If a user-defined copy constructor for myType is not provided, will myType‘s default copy constructor do a deep copy of ls?

Answer 1

Here ls is holding pointer to char. If copy constructor is not provided, will default copy constructor do the deep copy?

The default copy constructor will copy all members – i.e. call their respective copy constructors.¹ So yes, a std::vector (being nothing special as far as C++ is concerned) will be duly copied.

However, the memory pointed to by the char* elements inside the vector will of course not, since C++ doesn’t know and doesn’t care what the pointers point to.

But the solution here isn’t to provide a custom copy constructor. It’s to use a data structure instead of raw pointers (char*) which does. And this happens to be std::string (or std::vector<char> depending on the intent).

---

¹ Thus creating a transitive closure of the copy operation – this is how deep copying is generally implemented, but of course an implementor of the copy operation can always break out of it.